Dynamic Arrays and Amortized Analysis
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1 Department of Computer Science and Engineering Chinese University of Hong Kong
2 As mentioned earlier, one drawback of arrays is that their lengths are fixed. This makes it difficult when you want to use an array to store a set that may continuously grow and shrink with time. In this lecture, we will discuss clever tricks that allow us to design an array whose size can be varied efficiently! Our discussion also serves as a golden opportunity to introduce the method of amortized analysis.
3 Dynamic Array Problem Let S be a multi-set of integers that grows with time. At the beginning, S is empty. Over time, the integers of S are added by the following operation: insert(e): which adds an integer e into S. At any moment, let n be the number of elements in S. We want to store all the elements of S in an array A satisfying: 1 A has length O(n) 2 If an integer x was the i-th (i 1) inserted, then A[i] = x (i.e., x is at the i-th position of the array). This problem is dynamic, namely, the value of n continuously grows over time (initially, n = 0). The above requirements must be satisfied after every insertion.
4 Naive Algorithm Perform insert(e) as follows: If n = 0, then set n to 1. Initialize an array A of length 1, containing just e itself. Otherwise (i.e., n 1): Increase n by 1. Initialize an array A of length n. Copy all the n 1 elements of A over to A. Set A [n] = e. Destroy A, and replace it with A. This algorithm spends O(n) time on the n-th insertion. Altogether, it takes O(n 2 ) time to do n insertions.
5 We will improve the time of inserting n elements dramatically to O(n) (this is clearly optimal because every insertion must take at least constant time)! As a tradeoff, our array A may have a length up to 2n, which is still O(n).
6 A Better Algorithm We say that an array A is full, if the number of integers therein is already equal to its length. For example, if A was initialized with length 8, it is non-full if it has only up to 7 integers.
7 A Better Algorithm Perform insert(e) as follows: If n = 0, then set n to 1. Initialize an array A of length 2, containing just e itself. Otherwise (i.e., n 1), append e to A, and increase n by 1. If A is full, do the following Initialize an array A of length 2n. Copy all the n elements of A over to A. Destroy A, and replace it with A.
8 Example n = 1 n = 2 n = 3 n = 4 n = 5... n = 8
9 Analysis Cost of insertion when inserting the n-th element: If A is non-full after the insertion, O(1). Otherwise, O(n) the time incurred in expanding A. Suppose that the expansion time is at most cn, for some constant c > 0.
10 Analysis Array expansions happen infrequently: Initially, size 2. First expansion: size from 2 to 4. Second expansion: from 4 to The i-th expansion: from 2 i to 2 i+1. After n insertions, the size of A is at most 2n. Hence: 2 i+1 2n i log 2 n That is, there can be no more than log 2 n array expansions.
11 Analysis Therefore, the total cost of n insertions is bounded by: ( n ) log 2 n O(1) + c 2 i (1) i=1 where the first term corresponds to the compulsory constant time spent on each insertion, and the second term corresponds to the total cost of expanding. Formula (1) evaluates to O(n). i=1
12 Cleverer Analysis Next, we give an alternative analysis that proves the same conclusion with an elegant charging argument. Our algorithm maintains an invariant: After an array expansion, the new array has size 2n, namely, offering n empty positions.
13 Cleverer Analysis Suppose that an array expansion occurs at n, which takes c n time. The previous expansion happened at n/2. n/2 empty positions in the previous array. n/2 insertions have taken place since the previous expansion. Charge the c n cost over those n/2 insertions. c n Each insertion bears additional n/2 = 2c = O(1) cost. Therefore, the total cost of n insertions is O(n).
14 Example n = 1 n = 2 expanding cost charged on the 2nd element n = 3 n = 4 expanding cost charged on elements 3-4 n = 5... n = 8 expanding cost charged on elements 5-8
15 The Stack-with-Array Problem Let S be a multi-set of integers that grows with time. At the beginning, S is empty. We must support the following stack operations: push(e): which adds an integer e into S. pop: which removes from S the most recently inserted integer. At any moment, let m be the number of elements in S. We want to store all the elements of S in an array A satisfying: 1 A has length O(m) 2 A[1] is the least recently inserted element, A[2] the second least recently inserted,..., A[m] the most recently inserted. We will denote by n the number of operations processed so far.
16 The Stack-with-Array Problem We will give an algorithm for maintaining such an array by handling n operations in O(n) time, namely, each operation is processed in O(1) amortized time.
17 The Stack-with-Array Problem We say that 1 (Same as before) an array A is full, if the number of integers therein is equal to its length. 2 A is sparse if the number of integers therein is equal to 1/4 of its length (we will ensure that the length is a multiple of 4). We will stick to the invariant that, when an array is created, it is always half full, namely, if its size is s, it contains exactly s/2 elements.
18 Push Perform push(e) in the same way as an insertion in the dynamic array problem.
19 Pop Perform pop as follows: Return the last element of A, and decrease n by 1. If A is sparse, shrink the array as follows: Initialize an array A of length 2n. Copy all the n elements of A over to A. Destroy A, and replace it with A.
20 Example Next, we use the algorithm to perform 11 pushes and then 9 pops on an initially empty stack. n = 1, push n = 2, push... n = 4, push... n = 8, push... n = 11, push
21 Example... n = 17 pop n = 18, pop n = 19, pop n = 20, pop
22 Cost Analysis We will prove that our algorithm performs any sequence of n operations (each being either a push or a pop) using O(n) time in total, i.e., O(1) amortized time per operation. We will focus on the cost of array expanding and array shrinking collectively referred to as an overhaul. Applying a charging argument, we will charge the cost of an overhaul only on those operations that occurred between the last overhaul and the current one. In this way, we ensure that no operations are charged more than once. We will prove that each operation is charged at most O(1) cost, which indicates that the total cost of all operations is O(n).
23 Cost Analysis Array Expansion size s size 2s s/2 charge the cost of the array expansion on the push operations that entered these elements The cost of the expansion is at most c 1 s for some constant c 1. By charging the cost over the s/2 push operations as indicated above, each operation bears at most 2c 1 cost. Think: Why must these operations have occurred between the last overall and the current one?
24 Cost Analysis Array Shrinking recall that the array had s/2 elements when it was created. among them, s/4 have been removed. charge the cost of the shrinking on the pop operations that removed those s/4 elements. size s s/2 The cost of the expansion is at most c 2 s for some constant c 2. By charging the cost over the s/4 push operations as indicated above, each operation bears at most 4c 2 cost. Think: Why must these operations have occurred between the last overall and the current one?
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